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15 April, 01:06

Rule: Divide by 2, then add 6. 56, 100, 34, 23 A. 23, 34, 56, 100 B. 23, 56, 34, 100 C. 100, 34, 56, 23 D. 100, 56, 34, 23

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  1. 15 April, 01:08
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    For this problem, it is described that a set of number (x, y, z, ...) is calculated by a series of identical operation using the preceding number. In this problem if we let x as the first number, the second number y is, y = (x/2) + 6. Then the 3rd number z is equal to, z = (y/2) + 6, so on. Thus for this problem, the solution is D: 100, 56, 34, 23 through trial and error.
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